`" If " |{:(x^(2)+x,,x+1,,x-2),(2x^(2)+3x-1,,3x,,3x-3),(x^(2)+2x+3,,2x-1,,2x-1):}|=xA +B` then

Options

A. `|{:(1,,1,,1),(-1,,-3,,3),(4,,0,,0):}|`
B. `|{:(0,,1,,2),(1,,-2,,3),(-4,,0,,0):}|`
C. `|{:(1,,1,,-2),(-3,,-2,,3),(4,,0,,1):}|`
D. `|{:(0,,1,,-2),(-1,,-3,,3),(4,,0,,0):}|`

Answer

A, D

Explanation

Let `Delta= |{:(x^(2)+x,,x+1,,x-2),(2x^(2)+3x-1,,3x,,3x-3),(x^(2)+2x+3,,2x-1,,2x-1):}|`
Applying `R_(2) to R_(2) -2R_(1) " and " R_(3) to R_(3) -R_(1)` we get
`Delta = |{:(x^(2)+x,,x+1,,x-2),(x-1,,x-2,,x+1),(x+3,,x-2,,x+1):}|`
`=|{:(x^(2),,x+1,,x-2),(0,,x-2,,x+1),(0,,x-2,,x+1):}| + |{:(x,,x+1,,x-2),(x-1,,x-2,,x+1),(x+3,,x-2,,x+1):}|`
`=0 +|{:(x,,x+1,,x-2),(x-1,,x-2,,x+1),(x+3,,x-2,,x+1):}|`
Applying `R_(2) to R_(2)-R_(1)" and " R_(3) to R_(3)-R_(2)` we get
`Delta= |{:(x,,x+1,,x-2),(-1,,-3,,3),(4,,0,,0):}|`
`=|{:(x,,x,,x),(-1,,-3,,3),(4,,0,,0):}|+|{:(0,,-1,,-2),(-1,,-3,,3),(4,,0,,0):}|`
`=|{:(1,,1,,1),(-1,,-3,,3),(4,,0,,0):}|x + |{:(0,,1,,-2),(-1,,-3,,3),(4,,0,,0):}|`